0
\$\begingroup\$

I ordered several LMC6482IN opamp without reading the datasheet details! I want to measure high side current with mA accuracy, for that I need the opamp output to go all the way to ground 0V. in the datasheet it says:

So basically if supply voltage is 5V having a rail-to-rail output should be "guaranteed" and powering opamp with higher than that means we get some voltage swing (up to 20mV), right? but the output doesn't go below ~9mV!

  • Why opamp doesn't behave as datasheet "guaranteed"?
  • How can I achieve 0V output with this opamp?
  • If I connect opamp ground to -1V it would work?
\$\endgroup\$
  • 2
    \$\begingroup\$ This link to the datasheet is much better: ti.com/lit/ds/symlink/lmc6482.pdf \$\endgroup\$ – Bimpelrekkie Oct 16 at 8:56
  • 3
    \$\begingroup\$ Generally it's not possible for any rail-to-rail op amp to get all the way to within 0mV of its supply rails. Transistors can't have zero resistance... Getting within 20mV is really good dynamic range. Non-"rail-to-rail" op amps need even more headroom. \$\endgroup\$ – MarkU Oct 16 at 8:59
  • 1
    \$\begingroup\$ two inputs connected to two sides of 0.1 ohm resistor. Winny asked about the load, that means, what is connected to the opamp's output. You're only describing the input. \$\endgroup\$ – Bimpelrekkie Oct 16 at 9:05
  • 1
    \$\begingroup\$ So no feedback even, how on earth are you going to get that "with mA accuracy"? You will be measuring opamp offset multiplied by the open loop gain, i.e. you won't be able to measure *anything useful. You should include the schematic of your circuit. \$\endgroup\$ – Bimpelrekkie Oct 16 at 9:25
  • 1
    \$\begingroup\$ No you should not connect anything, you should study how to use opamps, look at other "high side current measuring" circuits, learn how they work. You now seem to be in a "connecting stuff but not having a clue" mode and that is wasting your time (and ours as well). We're assuming that you know what you're doing and that means that you would be using the opamp properly with feedback. You're not doing that so all bets are off as what you're doing it simply WRONG. \$\endgroup\$ – Bimpelrekkie Oct 16 at 9:32
1
\$\begingroup\$

Why opamp doesn't behave as datasheet "guaranteed"?

It does, but only the values in the tables are guaranteed. The rest is, well, up to anyone's interpretation.

How can I achieve 0V output with this opamp?

Maybe by loading it with only resistor to ground. 9 mV sounds quite good to me already. If you really need to go to 0 V, use a negative supply rail!

If I connect opamp ground to -1V it would work?

Yes, but don't call it "opamp ground" but call it "negative supply rail". A ground should have 0 V, you suggest -1 V so then it's no longer a ground.

\$\endgroup\$
  • \$\begingroup\$ just connecting -1V to negative supply rail will drag the output to zero? or output will be -991mV? \$\endgroup\$ – Hamed Oct 16 at 9:23
  • 1
    \$\begingroup\$ Hmm, that remark means that you need to learn a lot more about opamp circuits. Using a -1 V supply makes it possible for the opamp to drive its output to 0 V exactly. It could also drive it to -991 mV but that's pointless. The opamp should be in a feedback circuit so that the value at its output is well controlled. I'm suspecting that you're using the opamp in an open loop circuit and that is a big no-no for any circuit that needs to be accurate. \$\endgroup\$ – Bimpelrekkie Oct 16 at 9:28
  • 1
    \$\begingroup\$ Before asking more, please read the free Ebook "Opamps for Everyone": web.mit.edu/6.101/www/reference/op_amps_everyone.pdf \$\endgroup\$ – Bimpelrekkie Oct 16 at 9:28
  • \$\begingroup\$ True, i didn't study electronics and only have basic knowledge, as a hobbyist it's an advanced thing to do for me. but I'm trying and learning. thanks for the eBook link. \$\endgroup\$ – Hamed Oct 16 at 9:32
  • 1
    \$\begingroup\$ That's all fine, we've all been there :-) So do what I used to do when I was a rookie: I looked at and studied any circuit I could get my hands on and tried to understand how they work. Electronics easily gets too complicated to figure out on your own so learn from example, see what others do and do the same. \$\endgroup\$ – Bimpelrekkie Oct 16 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.